This is module 19 of Applications in Engineering Mechanics. We're going to continue on with the cable support system that, with concentrated loads that we started last time. We're going to be able to now find the tensile loads in each one of the cable sec, segments. And we're going to be able to determine the sag at particular points along the cable system with concentrated loads. And so here's the problem we started last time. We determined the reactions at points A and B, now we want to find the tension in each cable sec, section. And so let's go ahead and look at this section or segment from B to point P2. How would you find that that tension? What, what might you do? Okay, what you should have thought about doing was to go ahead and do a joint cut of a po, joint B. and if you draw that free body diagram, this is what it looks like. We found Bx and By last time. We have the tension in B2 which is unknown, but we do know it, it's slope. It's a four on six slope, and so you should now be able to solve for the tension in B2 directly. think about what equation of the equilibrium you'd use and go ahead, and solve for that and then come on back. 'Kay, the the equation that you could use is the sum of the forces in the y direction. if I substitute those in, I find that the tension in B2 is 2980 pounds. So we've, we've got that that section taken care of. Let's look at a, another option for finding TB2 and also finding Bx and Ax, the external reactions. Last time we did the external reactions using the method of sections. Now I'm going to use geometry. And so if I look at this joint B, I know that this By, we found that to be 1653.3 pounds. Bx was originally unknown and TB2 was originally unknown. What I can do is that I can now use similar triangles, so that this triangle, this side would be Bx, this side would be 1653. It's an analogous search similar to this triangle right here, so I can write that 1653.3 is related to the fourth side as Bx is related to the sixth side. And so I can solve for Bx directly. Bx ends up equaling 2480 pounds in tension, just like I found before. So I can use that method of geometry as an alternative. Now that I know Bx is equal to 2480. I know that this 2480 and 1653 pounds must be balanced by TB2. I can use Pythogoram's theorem to find out what the total magnitude is. You can do that on your own and then come on back. And oops, what you should find is that therefore TB2 equals the square root of B sub x squared plus 1653.3 squared. I found B sub x to be 2480. And so solving for TB2, we get 2980 pounds in tension. So now, I've taken care of both this section and this section. I, I'm sorry, just, just this section. Now we've got to go on and do the section from P1 to P2. We want to find the tension in that segment. And so my question is, how might you do that? the way I've done it is to go ahead and do a section cut. I'm going to actually look at this left hand side section. So draw a free body diagram of that. Okay, if you draw the free body diagram of that section, this is what it should look like. we know what Ax and Ay were, we solved for those as the reactions. We have this 500 pound force down. we don't know these reactions in T12 or do we? Do we know anything about T12 in the x direction. Well, what you should say is it's equal, it's got to balance to the 2480. So it's, it's 2480 pounds. And remember, the x component along any cut of the cable has to be this magnitude of 2480 pounds. So, now I should be able to find T12 directly using, first of all finding the y component by summing forces in the y direction. Do that on your own and then come on back. So you've got some of the forces in the y, I find T12 in the y direction is 246.7. Now I have both the x and y components. And I'll be able to find their the total result of T12 as being 2492 pounds in tension using Pythagoream's theorem. So now I've got this section and this section. I have one more section to do. How might you do that section. the section from A to P1. Think about it, and then let's do it together. Okay. We can do a joint cut. So we're going to cut this joint out. That will allow us to find A to P1. there's my joint cut, and you can sum forces in the y direction to find T1. TA1's y component or NX component. They've got to be the same and balance out the force, the external force reactions at A, so you can just use Pythagorean theorem to find TA1. The last part of the problem is to determine the sag at point P1. To do that, let's go ahead and again draw this section over here including the sag right P1 which is unknown. We'll use geometry to solve for the sag similar triangles again. If I look out here and do this triangle, this is the 746 side. this is 2480. And so I've got the 2480 side. These, these are both similar triangles, the 2480 side of this triangle corresponds to the 8 foot [INAUDIBLE] side of this triangle. And the 746.7 side of this triangle corresponds to the YP1 side which is my sag at 1. So I get, at P1, so I get YP1. If I solve that, is equal to 2.41 feet. And now I've solved A, B and C of the problem. And we're good to go.